Answer
$v_{f}\approx41 m/s$
Work Step by Step
We will use the formula $v^{2}_{f}=v^{2}_{0}+2gx$ to find the speed with which the package hits the ground ($v_{f}$).
We know that $v_{0}=12$, $g=9.8$ and $x=80$. Substituting these values in the formula and solving:
$v_{f}=\sqrt {v^{2}_{0}+2gx}=\sqrt {12^{2}+2(9.8)(80)}=41.4\approx41 m/s$
